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Chapter 27: Problem 29

An electron in a magnetic field moves counterclockwise on a circle in the \(x y\) -plane, with a cyclotron frequency of \(\omega=1.2 \cdot 10^{12} \mathrm{~Hz}\). What is the magnetic field, \(\vec{B}\) ?

Short Answer

Expert verified
Answer: The magnetic field acting on the electron is \(\vec{B} = 6.84 \times 10^{-4} \mathrm{T} \hat{z}\).

Step by step solution

01

Formula for cyclotron frequency

The formula for cyclotron frequency is given by: \(\omega = \frac{qB}{m}\) where \(\omega\) is the cyclotron frequency, \(q\) is the charge of the particle, \(B\) is the magnitude of the magnetic field, and \(m\) is the mass of the particle. For an electron, the charge \(q = -e\) and the mass \(m = m_e\), where \(e = 1.6 \times 10^{-19} \mathrm{C}\) and \(m_e = 9.11 \times 10^{-31} \mathrm{kg}\).
02

Rearranging the formula to find the magnitude of the magnetic field

Rearrange the formula to solve for the magnitude of the magnetic field \(B\): \(B = \frac{m\omega}{q}\)
03

Substitute the known values and calculate the magnetic field magnitude

Substitute the given value of \(\omega\) and known values of \(e\) and \(m_e\) into the formula and calculate \(B\): \(B = \frac{(9.11 \times 10^{-31} \mathrm{kg})(1.2 \times 10^{12} \mathrm{s}^{-1})}{1.6 \times 10^{-19} \mathrm{C}}\) After calculating, we get: \(B = 6.84 \times 10^{-4} \mathrm{T}\)
04

Determine the direction of the magnetic field

Here, as the electron moves in counterclockwise direction, the direction of the magnetic field \(\vec{B}\) must be perpendicular to the plane of its motion (xy-plane), along the positive z-direction. This follows the right-hand rule. So, the magnetic field \(\vec{B}\) is: \(\vec{B} = 6.84 \times 10^{-4} \mathrm{T} \hat{z}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field Calculation
When calculating the magnetic field for a moving electron, we start by using the concept of cyclotron frequency. The formula for cyclotron frequency is:
  • \( \omega = \frac{qB}{m} \)
This formula helps us understand the relationship between the charge of the particle, the mass of the particle, and the magnetic field. Here, \( \omega \) represents the cyclotron frequency, \( q \) is the charge, \( B \) is the magnitude of the magnetic field, and \( m \) is the mass of the particle.
For an electron, \( q = -e \) and \( m = m_e \), where \( e \) is the elementary charge \( (1.6 \times 10^{-19} \mathrm{C}) \) and the electron mass is \( m_e = 9.11 \times 10^{-31} \mathrm{kg} \).
To find the magnitude of the magnetic field, we rearrange the formula:
  • \( B = \frac{m\omega}{q} \)
The known value of \( \omega \) and the constants for \( e \) and \( m_e \) are substituted into this equation to calculate the magnetic field. This calculation gives us the field magnitude, \( B = 6.84 \times 10^{-4} \mathrm{T} \).
Electron Motion in Magnetic Field
Electrons exhibit fascinating behavior when they move through a magnetic field. They travel in a circular pattern due to the Lorentz force acting upon them, which is perpendicular to both the velocity of the electron and the direction of the magnetic field. This results in a circular motion, known as cyclotron motion.
For the given problem, we know the electron moves counterclockwise in the xy-plane. This circle is a result of the continuous perpendicular force that keeps changing the electron's direction without altering its speed.
  • The velocity vector of the electron is always tangent to the circle.
  • The magnetic field provides the inward centripetal force necessary for circular motion.
This type of motion is typical for charged particles in magnetic fields and is central to the operation of devices such as cyclotrons and magnetic confinement fusion devices.
Right Hand Rule
The Right Hand Rule is a simple yet powerful tool to determine the direction of magnetic fields. It is especially useful when considering the direction of forces in a magnetic field. To visualize the Right Hand Rule:
  • Point your thumb in the direction of the charge's velocity (here, you assume a positive charge for rule application).
  • Curl your fingers in the direction of the magnetic field.
  • Your thumb then points in the force direction exerted by the field on the charge.
Since our problem involves an electron, which has a negative charge, the force direction will actually flip. In this scenario of counterclockwise electron motion in the xy-plane, the magnetic field \( \vec{B} \) is along the positive z-direction based on the Right Hand Rule.
Therefore, using the Right Hand Rule allows us to confidently determine that the magnetic field points out of the plane, matching the solution鈥檚 final step. This fundamental rule is instrumental in solving many problems involving electromagnetism.

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Most popular questions from this chapter

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible? A blue problem number indicates a worked-out solution is available in the Student Solutions Manual. One \(\bullet\) and two \(\bullet\) indicate increasing level of problem difficulty.

The Earth is showered with particles from space known as muons. They have a charge identical to that of an electron but are many times heavier \(\left(m=1.88 \cdot 10^{-28} \mathrm{~kg}\right)\) Suppose a strong magnetic field is established in a lab \((B=0.50 \mathrm{~T})\) and a muon enters this field with a velocity of \(3.0 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) at a right angle to the field. What will be the radius of the resulting orbit of the muon?

An electron is moving with a constant velocity. When it enters an electric field that is perpendicular to its velocity, the electron will follow a ________ trajectory. When the electron enters a magnetic field that is perpendicular to its velocity, it will follow a ________ trajectory.

A helium leak detector uses a mass spectrometer to detect tiny leaks in a vacuum chamber. The chamber is evacuated with a vacuum pump and then sprayed with helium gas on the outside. If there is any leak, the helium molecules pass through the leak and into the chamber, whose volume is sampled by the leak detector. In the spectrometer, helium ions are accelerated and released into a tube, where their motion is perpendicular to an applied magnetic field, \(\vec{B},\) and they follow a circular orbit of radius \(r\) and then hit a detector. Estimate the velocity required if the orbital radius of the ions is to be no more than \(5 \mathrm{~cm},\) the magnetic field is \(0.15 \mathrm{~T}\) and the mass of a helium- 4 atom is about \(6.6 \cdot 10^{-27} \mathrm{~kg}\). Assume that each ion is singly ionized (has one electron less than the neutral atom). By what factor does the required velocity change if helium- 3 atoms, which have about \(\frac{3}{4}\) as much mass as helium- 4 atoms, are used?

The figure shows a schematic diagram of a simple mass spectrometer, consisting of a velocity selector and a particle detector and being used to separate singly ionized atoms \(\left(q=+e=1.60 \cdot 10^{-19} \mathrm{C}\right)\) of gold \((\mathrm{Au})\) and molybdenum (Mo). The electric field inside the velocity selector has magnitude \(E=1.789 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\) and points toward the top of the page, and the magnetic field has magnitude \(B_{1}=1.00 \mathrm{~T}\) and points out of the page. a) Draw the electric force vector, \(\vec{F}_{E},\) and the magnetic force vector, \(\vec{F}_{B},\) acting on the ions inside the velocity selector. b) Calculate the velocity, \(v_{0}\), of the ions that make it through the velocity selector (those that travel in a straight line). Does \(v_{0}\) depend on the type of ion (gold versus molybdenum), or is it the same for both types of ions? c) Write the equation for the radius of the semicircular path of an ion in the particle detector: \(R=R\left(m, v_{0}, q, B_{2}\right)\). d) The gold ions (represented by the black circles) exit the particle detector at a distance \(d_{2}=40.00 \mathrm{~cm}\) from the entrance slit, while the molybdenum ions (represented by the gray circles) exit the particle detector at a distance \(d_{1}=19.81 \mathrm{~cm}\) from the entrance slit. The mass of a gold ion is \(m_{\text {gold }}=\) \(3.27 \cdot 10^{-25}\) kg. Calculate the mass of a molybdenum ion.
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